Problem: Solve for $x$ : $7\sqrt{x} + 4 = 3\sqrt{x} + 8$
Subtract $3\sqrt{x}$ from both sides: $(7\sqrt{x} + 4) - 3\sqrt{x} = (3\sqrt{x} + 8) - 3\sqrt{x}$ $4\sqrt{x} + 4 = 8$ Subtract $4$ from both sides: $(4\sqrt{x} + 4) - 4 = 8 - 4$ $4\sqrt{x} = 4$ Divide both sides by $4$ $\frac{4\sqrt{x}}{4} = \frac{4}{4}$ Simplify. $\sqrt{x} = 1$ Square both sides. $\sqrt{x} \cdot \sqrt{x} = 1 \cdot 1$ $x = 1$